Download Q1 [20 points] Q2 [20 points]

BU-CmpE 220 Discrete Mathematics
bingol/Fall 2010
Midterm #2, 2 hours
Dec 07, 2010
Read me. Answer n should go to page n. Be clear, precise, short, necessary and sufficient in your answers.
Otherwise prepare to lose points. Open textbook, course handouts and your own handwritten notes, only. Keep the
question sheet. Show your own work. No questions. No exchange of information, books, notes, pencils etc. No
photocopied material. No phones. Good luck.
Q1 [20 points]
How many consecutive 0’s does the decimal number (base ten) 72! have on its right end?
Solution The number of 0’s on the right end depends on how many 10’s does 72! have as factor. Since
10 = 2 · 5, we need to find the number of 2’s and 5’s in 72! which is equal to the powers of 2 and 5 in the
prime factorization of 72!. Let 72! = 2p1 · 3p2 · 5p3 · 7p4 . . . be the prime factorization of 72!. Then the number
of 0’s on the right end would be min{p1 , p3 }. It could be shown that p1 > p3 . So we need to find p3 .
In the terms of multiplication 72! = 72 · 71 · . . . · 2 · 1 every other 5 terms has a 5 as prime factor such as
5, 10, 15. There are ⌊ 72
5 ⌋ = 14 such numbers.
Note that 14 counts 5’s as a single factor, that is 51 , but 25 and its multiples have 5 twice as factors,
that is 52 . There are ⌊ 72
25 ⌋ = 2 such numbers, namely 25 and 50.
Therefore the number of 0’s on the right end of 72! = p3 = 14 + 2 = 16.
Note that 125 and its multiples would have 5 as triple factors, that is 53 . Since 72 < 125 we do not count
those.
Q2 [20 points]
Let an denote the number of binary strings of lenght n that contain the string “01”.
a) Find a recurrence relation for an .
b) What are the initial conditions?
Solution.
a) Consider a binary string of length n − 1:
i) If this string contains the string “01” by appending 0 or 1 to its rigth end we obtain a string of
length n which contains “01”. So for every string of length n − 1 which contains “01”, we can obtain
two strings of length n which contains “01”.
|
{z
}
(n−1)-bit string which contains “01”
0| or
{z 1} =⇒ 2 ∗ an−1 such strings
nth bit
ii) If this string does not contain the string “01”, the only way that we can obtain a string of length n
which contains “01” is if its rightmost bit is a 0 and we append a 1 to its right end. Hence, for every
string of length n − 1 which does not contain “01” and whose right most bit is a 0, we have a string of
length n which contains “01”. The number of strings of length ”n-1” which does not contain “01” and
whose rightmost bit is “0” is equal to the number of (n − 2)-bit strings which do not contain “01”. We
can calculate this by subtracting the number (n − 2)-bit strings which contains “01” (an−2 ) from the
number of all strings of length n − 2 which is 2n−2 .
|
{z
}
(n−2)-bit string which does not contain “01”
0
|{z}
(n−1)th bit
1
1
|{z}
nth bit
=⇒ 2n−2 − an−2 such strings
an
=
=
2an−1 + 2n−2 − an−2 or in a better form
2an−1 − an−2 + 2n−2 for n ≥ 3.
b) The initial conditions are:
a1 = 0 since the strings of length 1, “0” and “1”, do not contain the string “01”.
a2 = 1 since among the strings of length 2, “00”, “01”, “10” and “11”, only “01” contains the string
“01”.
Q3 [20 points]
Show that for every positive integer n, there exists a positive multiple of n whose decimal expansion contains
only digits 0 and 7. (Hint: Use pigeon hole principle)
Solution. Consider the numbers:
7, 77, 777, 7777, ........, 777...777
| {z }
n+1 7′ s
i.e. the ith number contains i 7’s If we consider, the remainder that a number can have when divided to n
as our bins, there are n bins: 0, 1, 2..., n − 1.
G G G
G
.........
0
1
2
n−1
and the n + 1 numbers described above will be the pigeons. Since there are n + 1 numbers and n bins, there
will be a bin, say r (0 ≤ r ≤ n − 1), which contains at least two numbers. This means that there will be two
numbers, say i and j (i contains i 7’s and j contains j 7’s ), which have the same remainder when divided
by n.
⇒ i = nk + r and j = nl + r for some k, l ∈ N
⇒ i − j = nk − nl Let i > j without loss of generality
⇒ i − j = n(k − l)
⇒ i − j is a multiple of n.
Also, i − j contains only 0’s and 7’s, since it is of the form |77777
{z. . . 7} 0| . . . 000000
{z
}. As in the case of:
i−j
j
7777 . . . 7
77 . . . 7
7700 . . . 0.
So every positive integer n has a positive multiple which contains only 0’s and 7’s in its decimal expansion.
Q4 [20 points]
Prove the following theorem.
Theorem 0.1 (Cantor’s Theorem). Let A be a set. Then |A| =
6 |2A |.
Solution.
In other words “No set is the same size as its power set”.
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We show that f is not a surjection, therefore f is not a bijection.
Suppose f : A −→ 2A . Define B = {x ∈ A | x ∈
/ f (x)}. B ⊆ A. This means B ∈ 2A .
We claim that ∀x ∈ A f (x) 6= B, that is B ∈
/ f (A). If the claim is correct, that means f cannot map
into B. Hence f is not a surjection. Therefore f cannot be a bijection.
There is no bijection from A to 2A . Therefore |A| =
6 |2A |
Theorem 0.2. ∀x ∈ A f (x) 6= B, that is B ∈
/ f (A).
Proof. Suppose ∃b ∈ A f (b) = B. Ask if b ∈ B?
i) b ∈ B case: By definition of B, b ∈
/ f (b). Since f (b) = B, we have b ∈
/ B. Contradiction.
ii) b ∈
/ B case: Since f (b) = B, this means b ∈
/ f (b). This means b ∈ B since B is defined so. Contradiction.
So we obtain b ∈ B ⇔ b ∈
/ B. This contradiction means ¬∃b ∈ A f (b) = B. That is ∀b ∈ A f (b) 6= B.
The theorem is proved.
Q5 [20 points]
Let G = [A, ∗] be a group and G be the set of subgroups of G. Define the relation < on the set G as follows:
∆
G1 < G2 ←→ G1 is a subgroup of G2 where G1 = [A1 , ∗1 ], G2 = [A2 , ∗2 ] ∈ G. Answer the following
questions on < by providing a proof or a counter example. (Use the notation Gi = [Ai , ∗i ] for the elements
of G).
a) Is < reflexive?
b) Is < symmetric?
c) Is < antisymmetric?
d) Is < transitive?
e) Is < an equivalence relation?
f) Is < a partial ordering?
Solution.
a) Yes, < is reflexive since a group is a trivial subgroup of itself. Let G1 = [A1 , ∗1 ]. Then G1 < G1 , since
i) G1 is a group itself.
ii) A1 ⊆ A1 , every set is a subset of itself.
iii) ∗1 is a restriction of itself by the definition of restriction.
Hence, all conditions in Definition 2.5 (Algebraic structures lecture notes) is satisfied and G1 < G1 is
true for every G1 ∈ G. Therefore, < is reflexive.
b) No, < is not symmetric. A counter example. Let G = [R, +]. Both G1 = [Z, +] and G2 = [{0}, +] are
subgroups of G and G1 < G2 . But G2 < G1 is not true since Z 6⊆ {0}
c) Yes, < is antisymmetric. Let G1 = [A1 , ∗1 ], G2 = [A2 , ∗2 ] ∈ G with G1 < G2 and G2 < G1 .
G1 < G2 ⇒ (A1 ⊆ A2 ) ∧ (∗1 is the restriction of ∗2 on A1 ).
G2 < G1 ⇒ (A2 ⊆ A1 ) ∧ (∗2 is the restriction of ∗1 on A2 ).
A1 ⊆ A2 ∧ A2 ⊆ A1 ⇒ A1 = A2 .
(∗1 is the restriction of ∗2 on A2 ) ∧ (∗2 is the restriction of ∗1 on A2 ) ∧ (A1 = A2 ) ⇒ ∗1 = ∗2
Hence, G1 = G2 . Therefore, < is antisymmetric.
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d) Yes, < is transitive. Let G1 = [A1 , ∗1 ], G2 = [A2 , ∗2 ], G3 = [A3 , ∗3 ] ∈ G with G1 < G2 and G2 < G3 .
G1 < G2 ⇒ (A1 ⊆ A2 ) ∧ (∗1 is the restriction of ∗2 on A1 ).
G2 < G3 ⇒ (A2 ⊆ A3 ) ∧ (∗2 is the restriction of ∗3 on A2 ).
(A1 ⊆ A2 ∧ A2 ⊆ A3 ) ⇒ A1 ⊆ A3 .
(∗1 is the restriction of ∗2 on A1 ) ∧ (∗2 is the restriction of ∗3 on A2 ) ⇒ (∗1 is the restriction of ∗3 on
A1 ).
( A1 ⊆ A3 ) ∧ (∗1 is the restriction of ∗3 on A1 ) ∧ (G1 and G2 are groups ) ⇒ G1 < G3 . Therefore, <
is transitive.
e) No, < is not an equivalance relation since it is not symmetric.
f) Yes, < is a partial ordering since it is reflexive, antisymmetric and transitive as shown above.
Quotation of the day:
David Hilbert: “ The infinite! No other question has ever moved so profoundly the spirit of man.”
(source: http://en.wikipedia.org/wiki/Hilbert)
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